Let x be the distance (in meters) along the road that the car has traveled andh be the distance (in meters) between the car and the observer.
Part (a):
Before the car passes the observer, we have [tex] \frac{dh}{dt} \ \textless \ 0[/tex]; after it passes, we have [tex]\frac{dh}{dt} \ \textgreater \ 0[/tex].
So at the instant it passes we have [tex]\frac{dh}{dt} = 0[/tex], given that [tex]\frac{dh}{dt} = 0[/tex] varies continuously since the car travels at a constant velocity.
Therefore, the rate the distance from the observer to the car increasing when the car passes in front of the observer is 0m/s
Part (b):
By the Pythagorean Theorem, we have
[tex]h^2=x^2+100^2[/tex]
Thus,
[tex]2h\frac{dh}{dt}=2x\frac{dx}{dt} \\ \\ \Rightarrow\frac{dh}{dt}= \frac{x}{h} \frac{dx}{dt}[/tex]
where: [tex]\frac{dx}{dt}[/tex] is the velocity at which the car is travelling.
Since the car travels at 25 m/s, so after 10 seconds, the distance (in meters) along the road that the car has traveled, x, is given by
[tex]x=speed\times time=25\times10=250 \ meters.[/tex]
and
[tex]h^2=250^2+100^2=62,500+10,000=72,500 \\ \\ \Rightarrow h= \sqrt{72,500} =269.26 \ m[/tex]
Therefore,
[tex]\frac{dh}{dt}= \frac{x}{h} \frac{dx}{dt} \\ \\ = \frac{250}{269.26} (25) \\ \\ =0.9285(25) \\ \\ =23.212 \ m/s[/tex]
